I need some help with an analysis problem.

Define $g:[-1,1]\to \Bbb R$ by $g(x)=(-1)^k/k^2$ for $|x|\in(1/(k+1),1/k], k=1,2,...$ and $g(0)=0$. Decide whether or not $g$ is differentiable at $0$ and prove your answer.

What I have:

When $k=1, |x|\in (1/2,1]$ and if $k=2,|x|\in(1/3,1/2]$ so if $k\to \infty, |x|\in(0,0]$

But how would I find if this is differentiable at zero or not?

  • $\begingroup$ Have you tried graphing this function? $\endgroup$ – Matthew Leingang Feb 19 '19 at 0:47
  • $\begingroup$ I'm not quite sure how I would graph it. $\endgroup$ – geoplanted Feb 19 '19 at 0:50
  • $\begingroup$ Good, so I think understanding what the function is doing well enough to graph it is a key step before deciding if it's differentiable at zero. What is $g(0.1)$? $g(0.01)$? $g(-0.1)$? Plot some points and get acquainted with $g$. $\endgroup$ – Matthew Leingang Feb 19 '19 at 0:52
  • $\begingroup$ Thats something I am struggling with. If g is a function of x, how would I graph it since everything is defined in terms of k? $\endgroup$ – geoplanted Feb 19 '19 at 0:57
  • 1
    $\begingroup$ From my understanding $g$ looks like a bunch of lines above and below the $x$ axis. The lines are getting shorter as they approach the origin. They also are getting closer together. For every $k$ there are two lines. The function is symmetrical. $\endgroup$ – ty. Feb 19 '19 at 1:05

We can look for the derivative by definition: $$g'(0) = \lim\limits_{h\to0}\frac{g(h)-g(0)}{h} = \lim\limits_{h\to0}\frac{g(h)}{h}$$

However, the way the function is defined is not very convenient for us to find the limit, because it is defined in terms of an arbitrary number $k$. That's why we will try to define the function in terms of $x$ only. Note that the condition $$|x| \in \left( \frac{1}{k+1}, \frac{1}{k} \right]$$ after playing around with inequalities (with $x\neq 0$), is equivalent with the condition $$\frac{1}{|x|} \in [k, k+1)$$ How can we express the function $g(x)=(-1)^k/k^2$ in terms $x$ only?

We can use the floor function. Note that the function $\lfloor x \rfloor$ is defined as the largest integer not less than $x$. With that in mind, it is easy to see that $k=\lfloor 1/|x| \rfloor$. Now the function can be written as $$g(x)=\frac{(-1)^{\lfloor 1/|x| \rfloor}}{\lfloor 1/|x| \rfloor^2}$$

Let's get back to the limit. Looking at the form of $g(h)/h$, we can intuit that this limit is zero. We shall prove this by definition. Namely, we must show that for all $\epsilon>0$, there exists $\delta>0$ such that $$0<|h|<\delta \implies \left|\frac{g(h)}{h}\right|<\epsilon$$

Since $$\left\lfloor \frac{1}{|h|} \right\rfloor \geq \frac{1}{2|h|} \implies \left\lfloor \frac{1}{|h|} \right\rfloor^2 \geq \frac{1}{4h^2}$$ for all sufficiently small $h$ (say for $|h|<\eta$), we have that

$$\left|\frac{g(h)}{h}\right| = \frac{1}{|h|\cdot\lfloor 1/|h| \rfloor^2} \leq \frac{4h^2}{|h|}=4|h|$$ For this to be smaller than $\epsilon$, it suffices to choose $\delta<\min\{\epsilon/4,\eta\}$.

Thus, we have proven that $$g'(0)=\lim\limits_{h \to 0} \frac{g(h)}{h}=0$$ and $g$ is differentiable at the point $x=0$. $\ \ \rule{5pt}{5pt}$

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  • $\begingroup$ Let me know if you need any more clarification. $\endgroup$ – Haris Gušić Feb 19 '19 at 2:29

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